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php regex: lookbehind and lookahead and greediness problem

Answer 1

+2 A:

This should do it:

(?<=\[p2\]).+?(?=\[\/p2\])

I added the question mark to make the quantifier non-greedy.

Franz 2009-11-13 21:07:54

thanks for the answer

Jared Henderson 2009-11-13 21:14:19

Alternatively, you can specify the `U` modifier to indicate that all quantifiers in the pattern should be considered as non-greedy, e.g. `preg_match('/(?<=\[p2\]).+(?=\[\/p2\])/U', ....` Notice the **capital** `U` character will automatically convert the `+` sign to non-greedy, and a trailing `?` (greediness operator) would cause it to exhibit the behavior you are describing. Worth knowing.See: http://us2.php.net/manual/en/reference.pcre.pattern.modifiers.php

Dereleased 2009-11-13 21:25:11

@Dereleased: I don't quite understand. What exactly is the difference? Thanks for the comment, though.

Franz 2009-11-13 21:32:08

`/U` is overkill in this case, what with there being only one quantifier. But I would avoid using `/U` in any case; it makes your regex less portable (few regex flavors support it) and less readable. People expect quantifiers to be greedy, and have to make a conscious effort to override that expectation. That's a lot easier to do when you're looking at the `?` right next to the quantifier.

Alan Moore 2009-11-14 01:02:36

Answer 2

A:

Instead of using a regex modifier, you can use the standard perl style match modifier and add a ? after + or * to tell that specific portion to be non-greedy. Mentioned above, but specificity can help.

kbenson 2009-11-13 21:19:59

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php regex: lookbehind and lookahead and greediness problem

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